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3 years ago

Placement.

Physics

1 answer:

stealth61 [152]3 years ago

7 0

Answer:

kinetic energy will be equal to 0

Explanation:

this is because at final position velocity of body will become zero.

kinetic energy eill be 8 times

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horsena [70]

It will be cloudy and there will be rain.

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2 years ago

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What is the speed of sound

Charra [1.4K]

Answer:it is 343 meter per second in air.

Explanation:

it is 5 times more in liquids than in gases and 15 times more in solids than in gases.In air it is affected by various physical conditions such as pressure,humidity and temperature.

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2 years ago

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Misha Larkins [42]

Hi there!

The maximum deformation of the bumper will occur when the car is temporarily at rest after the collision. We can use the work-energy theorem to solve.

Initially, we only have kinetic energy:

$KE = \frac{1}{2}mv^2$

KE = Kinetic Energy (J)
m = mass (1060 kg)
v = velocity (14.6 m/s)

Once the car is at rest and the bumper is deformed to the maximum, we only have spring-potential energy:

$U_s = \frac{1}{2}kx^2$

k = Spring Constant (1.14 × 10⁷ N/m)

x = compressed distance of bumper (? m)

Since energy is conserved:

$E_I = E_f\\\\KE = U_s\\\\\frac{1}{2}mv^2 = \frac{1}{2}kx^2$

We can simplify and solve for 'x'.

$mv^2 = kx^2\\\\x = \sqrt{\frac{mv^2}{k}}$

Plug in the givens and solve.

$x = \sqrt{\frac{(1060)(14.6^2)}{(1.14*10^7)}} = \boxed{0.0198 m}$

3 0

1 year ago

Four traveling waves are described by the following equations, where all quantities are measured in SI units and y represents th

agasfer [191]

Answer:

$T_1=T_3=\dfrac{2\pi}{21}$

$T_2=T_4=\dfrac{2\pi}{42}$

Explanation:

Wave 1, $y_1=0.12\ cos(3x-21t)$

Wave 2, $y_2=0.15\ sin(6x+42t)$

Wave 3, $y_3=0.13\ cos(6x+21t)$

Wave 4, $y_4=-0.27\ sin(3x-42t)$

The general equation of travelling wave is given by :

$y=A\ cos(kx\pm \omega t)$

The value of $\omega$ will remain the same if we take phase difference into account.

For first wave,

$\omega_1=21$

$\dfrac{2\pi }{T_1}=21$

$T_1=\dfrac{2\pi}{21}$

For second wave,

$\omega_2=42$

$\dfrac{2\pi }{T_2}=42$

$T_2=\dfrac{2\pi}{42}$

For the third wave,

$\omega_3=21$

$\dfrac{2\pi }{T_3}=21$

$T_3=\dfrac{2\pi}{21}$

For the fourth wave,

$\omega_4=42$

$\dfrac{2\pi }{T_4}=42$

$T_4=\dfrac{2\pi}{42}$

It is clear from above calculations that waves 1 and 3 have same time period. Also, wave 2 and 4 have same time period. Hence, this is the required solution.

3 0

3 years ago

Can matter exist as a mixture of two or three phases? five an example

tensa zangetsu [6.8K]

Yes they can.

Don't really know any examples

8 0

3 years ago